The Complete Expectation of Life - Infinity is Really Big

Studying for Exam LTAM, Part 1.3

What is your life expectancy? Are you a smoker? A heavy drinker? Male or Female? And how well do you eat and exercise?

These factors can all affect how long you can “expect” to live.

However, the word “expect” can be misleading in this context. It is an average. It is not set in stone. In fact, it might not even be “typical”. There can be a lot of variation. In other words, for all the people in the world who are similar to you, there will be a wide variety of ages that they will live to.

Mathematically, your life expectancy is the expected value, or mean (average), of a continuous survival random variable that models the lives of people who are similar to you.

Expected Value and Complete Expectation of Life

Let $X$ be an arbitrary continuous random variable with probability density function (PDF) $f(x)$ . Let $H$ be an arbitrary “nice” function (such as a continuous function). The expected value of the “transformed” random variable $H(X)$ is $E[H(X)]=\displaystyle\int_{-\infty}^{\infty}h(x)f(x)\, dx$ .

When $H(X)=X$ , the answer is also denoted by $\mu$ and it is called the mean, or expected value, of $X$ . The equation in this case becomes $\mu=E[X]=\displaystyle\int_{-\infty}^{\infty}xf(x)\, dx$ . When $n$ is a positive integer and $H(X)=X^{n},$ then $E[H(X)]=E[X^{n}]$ is called the nth moment of $X$ .

Recall, however, that when $T$ is a continuous survival random variable, we can take our lower limit of integration to be zero, rather than $-\infty$ .

Let us return to the setting of actuarial science and focus on the standard notation of that subject. Let $T_{0}$ be the continuous lifetime of a new life (such as a newborn baby). For any $x>0$ , assuming survival to age $x$ , let $T_{x}=T_{0}-x$ be the corresponding remaining-life random variable. The mean of $T_{x}$ is denoted by $\stackrel{\circ}e_{x}$ . In other words, $\stackrel{\circ}e_{x}=\displaystyle\int_{0}^{\infty}xf_{x}(t)\, dt$ , where $f_{x}(t)$ is the PDF of $T_{x}$ .

The quantity $\stackrel{\circ}e_{x}$ is called the complete expectation of life for $(x)$ (recall that the symbol $(x)$ represents the individual of current age $x$ ). The word “complete” is used here because we are not just rounding down to their nearest whole-number age. We are including fractional ages.

The Survival Function and the Complete Expectation of Life

Let $T$ be an arbitrary continuous survival random variable with probability density function (PDF) $f(t)$ and cumulative distribution function (CDF) $F(t)=P[T\leq t]=\displaystyle\int_{0}^{t}f(s)\, ds$ . The corresponding survival function (SF) is $S(t)=P[T>t]=1-F(t)=1-\displaystyle\int_{0}^{t}f(s)\, ds=\displaystyle\int_{t}^{\infty}f(s)\, ds$ . Note that $f(t)=F'(t)=-S'(t)$ .

Let us make the additional (reasonable) assumption that $\displaystyle\lim_{t\rightarrow \infty}tS(t)=0$ . If we set $u=t$ and $dv=f(t)\, dt$ , then we can write $du=dt$ and $v=-S(t).$ From this, the formula for integration-by-parts can be used to write $\mu=E[T]=\displaystyle\int_{0}^{\infty}tf(t)\, dt=-\displaystyle\lim_{b\rightarrow \infty}(tS(t))|_{t=0}^{t=b}+\displaystyle\int_{0}^{\infty}S(t)\, dt=\displaystyle\int_{0}^{\infty}S(t)\, dt$ .

Recall from the previous post “The Force of Mortality (Hazard Rate Function)” that the notation for the survival function of $T_{x}$ , as a function of $t$ , is $\,_{t}p_{x}$ . Hence, we can now say that, when $\displaystyle\lim_{t\rightarrow \infty}t\cdot \,_{t}p_{x}=0$ , we have $\stackrel{\circ}e_{x}=\displaystyle\int_{0}^{\infty}\,_{t}p_{x}\, dt$ . This is the most common way that a mean is calculated in this context.

Uniform Distribution (De Moivre’s Law)

We now consider the same example as in the previous couple posts. We assume that $T_{0}$ has a uniform distribution over an interval $[0,\omega]$ , which results in $T_{x}$ having a uniform distribution over $[0,\omega-x]$ .

Technically, all the relevant functions are piecewise-defined for this example. However, we will just restrict ourselves to the most important intervals and not worry about this fact. The relevant formulae are as follows. The PDF is $f_{x}(t)=\frac{1}{\omega-x}$ for $0\leq t\leq \omega-x$ . The CDF is $F_{x}(t)=\frac{t}{\omega-x}$ for $0\leq t\leq \omega-x$ . And the SF is $\,_{t}p_{x}=1-\frac{t}{\omega-x}=\frac{\omega-x-t}{\omega-x}$ for $0\leq t\leq \omega-x$ .

We compute $\stackrel{\circ}e_{x}$ in two ways to see that we get the same answer.

First, $\stackrel{\circ}e_{x}=\displaystyle\int_{0}^{\infty}tf_{x}(t)\, dt=\displaystyle\int_{0}^{\omega-x}\frac{t}{\omega-x}\, dt=\frac{t^{2}}{2(\omega-x)}\Bigr\rvert_{t=0}^{t=\omega-x}=\frac{\omega-x}{2}$ .

Second, $\stackrel{\circ}e_{x}=\displaystyle\int_{0}^{\infty}\,_{t}p_{x}\, dt=\displaystyle\int_{0}^{\omega-x}\left(1-\frac{t}{\omega-x}\right)\, dt$ . From this we get the same answer as above: $\stackrel{\circ}e_{x}=\left(t-\frac{t^{2}}{2(\omega-x)}\right)\Bigr\rvert_{t=0}^{t=\omega-x}=\omega-x-\frac{\omega-x}{2}=\frac{\omega-x}{2}$ .

This should make intuitive sense. In the model, $\omega$ is the maximum age a newborn can attain before death. Therefore, if $(x)$ is at age $x$ , the maximum remaining life is $\omega-x$ . Since the distribution is uniform, the average remaining life is half of this, or $\frac{\omega-x}{2}$ .

The Complete Expectation of Life is a Function of the Attained Age

Conceptually-speaking, we approached these calculations as if $x$ was fixed. However, now that we are done, we are free to treat $x$ as a variable if we desire. Indeed, $\stackrel{\circ}e_{x}$ is a function of $x$ that can be graphed.

The animation below shows, for the uniform distribution example with $\omega=100$ , the graph of $\stackrel{\circ}e_{x}$ on the left and the graph of $\,_{t}p_{x}$ on the right (including when $t>\omega-x$ ). The slider tracks the value of $x$ from 0 to 99. The black dots show the relevant point on each graph (noticed the labeled coordinates).

The graphs of $\stackrel{\circ}e_{x}$ and $\,_{t}p_{x}$ with a varying black dot as the attained age $x$ increases from 0 to 99. Note that the second coordinate of the black dot for the graph on the right is 0.5 because the mean and median of a uniform distribution are equal.

The Mathematica code to create this animation is shown below.